//编写一个程序，通过填充空格来解决数独问题。 
//
// 数独的解法需 遵循如下规则： 
//
// 
// 数字 1-9 在每一行只能出现一次。 
// 数字 1-9 在每一列只能出现一次。 
// 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图） 
// 
//
// 数独部分空格内已填入了数字，空白格用 '.' 表示。 
//
// 
//
// 
// 
// 
// 示例 1： 
// 
// 
//输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".
//",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".
//","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6
//"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[
//".",".",".",".","8",".",".","7","9"]]
//输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8
//"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],[
//"4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9",
//"6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4",
//"5","2","8","6","1","7","9"]]
//解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
// 
// 
// 
// 
//
//
//
// 
//
// 提示： 
//
// 
// board.length == 9 
// board[i].length == 9 
// board[i][j] 是一位数字或者 '.' 
// 题目数据 保证 输入数独仅有一个解 
// 
//
// Related Topics 数组 哈希表 回溯 矩阵 👍 1676 👎 0


package leetcode.editor.cn;

// [37]解数独

public class SudokuSolver_37 {
    public static void main(String[] args) {
        Solution solution = new SudokuSolver_37().new Solution();
        solution.solveSudoku(new char[][]{
                {'5', '3', '.', '.', '7', '.', '.', '.', '.'},
                {'6', '.', '.', '1', '9', '5', '.', '.', '.'},
                {'.', '9', '8', '.', '.', '.', '.', '6', '.'},
                {'8', '.', '.', '.', '6', '.', '.', '.', '3'},
                {'4', '.', '.', '8', '.', '3', '.', '.', '1'},
                {'7', '.', '.', '.', '2', '.', '.', '.', '6'},
                {'.', '6', '.', '.', '.', '.', '2', '8', '.'},
                {'.', '.', '.', '4', '1', '9', '.', '.', '5'},
                {'.', '.', '.', '.', '8', '.', '.', '7', '9'}});
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {

        boolean flag = false;
        int[][] check = new int[3][9];

        public void solveSudoku(char[][] board) {
            makeCheck(board, check);
            process(board, 0);
        }

        private void makeCheck(char[][] board, int[][] check) {
            for (int x = 0; x < 81; x++) {
                int i = x / 9;
                int j = x % 9;
                int k = i / 3 * 3 + j / 3;
                if (board[i][j] != '.') {
                    changeBoard(check, i, j, k, board[i][j] - '0');
                }
            }
        }

        private void process(char[][] board, int x) {
            if (x == 81) {
                flag = true;
            }
            if (flag) {
                return;
            }
            int i = x / 9;
            int j = x % 9;
            int k = i / 3 * 3 + j / 3;
            if (board[i][j] != '.') {
                process(board, x + 1);
                return;
            }
            for (int l = 1; l <= 9; l++) {
                if (isValid(check, i, j, k, l)) {
                    changeBoard(check, i, j, k, l);
                    board[i][j] = (char) (l + '0');
                    process(board, x + 1);
                    if (flag) {
                        return;
                    }
                    board[i][j] = '.';
                    changeBoard(check, i, j, k, l);
                }
            }
        }

        private void changeBoard(int[][] check, int i, int j, int k, int n) {
            check[0][i] ^= 1 << (n - 1);
            check[1][j] ^= 1 << (n - 1);
            check[2][k] ^= 1 << (n - 1);
        }

        private boolean isValid(int[][] check, int i, int j, int k, int n) {
            if ((check[0][i] & (1 << (n - 1))) != 0) {
                return false;
            }
            if ((check[1][j] & (1 << (n - 1))) != 0) {
                return false;
            }
            return (check[2][k] & (1 << (n - 1))) == 0;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}